Answer
The required solution is $\frac{{}_{20}{{P}_{2}}}{2!}-{}_{20}{{C}_{2}}=0$.
Work Step by Step
The permutation and combination formulas respectively are:
${}_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ And ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$
Applying these formulas into the provided expression, we get:
$\begin{align}
& \frac{{}_{20}{{P}_{2}}}{2!}-{}_{20}{{C}_{2}}=\frac{20!}{2!\left( 20-2 \right)!}-\frac{20!}{2!\left( 20-2 \right)!} \\
& =0
\end{align}$
$\frac{{}_{20}{{P}_{2}}}{2!}-{}_{20}{{C}_{2}}=0$
